the weight of the child is 250 N. the height of the slide is 7m. the work done against friction as the child travels down the slide is 1300 J. what is the change in gravitational potential energy and the final kinetic energy of the child.
Answers
Answer:
A student applies a force to a cart to pull it up an inclined plane at constant speed during a physics lab. A force of 20.8 N is applied parallel to the incline to lift a 3.00-kg loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work done upon the cart and the subsequent potential energy change of the cart. PSYW
Answer: 13.2 J
There are two methods of solving this problem. The first method involves using the equation
W = F*d*cos(Theta)
where F=20.8 N, d=0.636 m, and Theta=0 degrees. (The angle theta represents the angle betwee the force and the displacement vector; since the force is applied parallel to the incline, the angle is zero.) Substituting and solving yields
W = F*d*cos(Theta) = (20.8 N)*(0.636 m)*cos(0) = 13.2 J.
The second method is to recognize that the work done in pulling the cart along the incline at constant speed changes the potential energy of the cart. The work done equals the potential energy change. Thus,
W=Delta PE = m*g*(delta h) = (3.00 kg)*(9.8 m/s/s)*(0.45 m) = 13.2 J
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